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4PM1/01 Jun 2024, Question 7: Tangents, Normals and Stationary Points (Further)

Pearson Edexcel International GCSE Further Pure Mathematics 4PM1 · Paper 1 · Calculator · 13 marks

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The question, in short

Use the curve y = x²/4 − 3√x + 8 and its sketch to verify a point, derive a normal line, and integrate for an area.

Summarised in our own words. Use the official paper for the exact wording and the diagram.

Part (a) · 1 mark

Verify the y-coordinate at x = 4 on the given curve equals 6.

  1. 1a = 4²/4 − 3√4 + 8
  2. 2= 4 − 6 + 8 = 6

Answera = 6 shown

Part (b) · 6 marks

Derive the normal line equation at (4, 6), showing it rearranges to 5y + 4x − 46 = 0.

  1. 1dy/dx = x/2 − (3/2)x^(−1/2)
  2. 2At x = 4: dy/dx = 2 − 3/4 = 5/4
  3. 3Gradient of normal = −4/5
  4. 4y − 6 = −(4/5)(x − 4)
  5. 55y − 30 = −4x + 16, so 5y + 4x − 46 = 0

Answer5y + 4x − 46 = 0 shown

Part (c) · 6 marks

Integrate to compute the exact area of the region bounded by the curve, normal, x-axis and x = 1.

  1. 1Area under C from 1 to 4 = ∫ from 1 to 4 of (x²/4 − 3√x + 8) dx
  2. 2= [x³/12 − 2x^(3/2) + 8x] from 1 to 4 = 61/4
  3. 3L meets the x-axis where 4x = 46, x = 11.5
  4. 4Triangle under L from x = 4 to 11.5: ½ × 7.5 × 6 = 45/2
  5. 5Area of R = 61/4 + 45/2 = 151/4

Answer151/4

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